A) \[{{K}_{2}}={{K}_{1}}\]
B) \[{{K}_{2}}=\frac{{{K}_{1}}}{2}\]
C) \[{{K}_{2}}=2{{K}_{1}}\]
D) \[{{K}_{2}}=\frac{{{K}_{1}}}{4}\]
Correct Answer: B
Solution :
| (b): \[{{K}_{1}}=\frac{1}{2}m\,{{v}^{2}}_{\max }=\frac{1}{2}m{{A}^{2}}\omega _{1}^{2}\] .(i) |
| \[{{K}_{2}}=\frac{1}{2}m\,A_{2}^{2}\omega _{2}^{2}\] .(2) |
| Here,\[{{A}_{2}}=A,\] |
| From eqn. (i) and (ii) |
| \[\frac{{{K}_{2}}}{{{K}_{1}}}=\frac{\omega _{2}^{2}}{\omega _{1}^{2}}\] \[\left( \omega =\sqrt{\frac{g}{l}} \right)\] |
| \[\frac{{{K}_{2}}}{{{K}_{1}}}=\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{l}_{1}}}{2{{l}_{1}}}\Rightarrow {{K}_{2}}=\frac{{{K}_{1}}}{2}\] |
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